help me solve the attached question.

2. Originally Posted by mathkidx
help me solve the attached question.
Do you know these?
1) The measure of an inscribed angle in a circle is equal to half the measure of the intercepted arc.
2) The measure of the angle made by a tangent line and a secant is half the meaure of the intercepted arc.
3) The base angles of an isosceles triangle are congruent or equal in measures.
4) The 3 interior angles of a triangle add up to 180 degrees.
5) The opposite interior angles of a cyclic quadrilateral are supplementary, or add up to 180 degrees.

If you do, then let us go to the diagram.

If KN = KL, then triangle NKL is isosceles. Then the base angles KNL and KLN are equal. Let us call each as angle theta or t.

Then, minor arc NK = 2(t).
Also, minor arc KL = 2(t).

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To solve for angle KLN:

angle PNL = (1/2)[minor arc NK + minor arc KL) = (1/2)(2t +2t) = 2t
angle PNL = 24deg +t .......as shown on the diagram
so,
2t = 24deg +t
t = 24 degrees -------**

So, since angle KLN is t, then angle KLN = 24 degrees. -----answer.

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To solve for angle NKL:

angle NKL = 180deg -t -t
angle NKL = 180 -24 -24 = 132 degrees -------answer.

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To solve for angle LMN:

Angle LMN is the supplement of angle NKL. so,
angle LMN = 180 -132 = 48 degrees. ------------answer.

We can check that in the diagram,
angle LMN = (1/2)[minor arc NK + minor arc KL)
angle LMN = (1/2)(2t +2t) = 2t == 2(24) = 48 degrees.